From Eisenbud's Commutative Algebra with A View Toward Algebraic Geometry (Theorem 17.4):
Let $M$ be a finitely generated $R$-module, where $R$ is Noetherian. If $$r= \min \{i : H^i(M\otimes K(x_1,\dots, x_n) ) \neq 0 \}, $$ then every maximal $M$-sequence in $I:=(x_1,\dots, x_n) \subseteq R$ has length $r$.
The proof in the text is as follows:
Let $y_1,\dots, y_s$ be a maximal $M$-sequence in $I$. By functoriality of the Koszul complex, $$ H^{*}(M\otimes K(x_1,\dots, x_n,y_1,\dots, y_s)) \cong H^*(M\otimes K(x_1,\dots, x_n)) \otimes \wedge R^s $$ so that $r$ is also the smallest integer for which $H^{i}(M\otimes K(x_1,\dots, x_n,y_1,\dots, y_s)) \neq 0$. Since the cohomology does not vanish on some dimension, we cannot have $IM=M$. Using Corollary 17.12, $s=r$ as desired. $\square$
Here is Corollary 17.12:
Suppose $x_1,\dots, x_i$ is a maximal $M$-sequence in $I=(x_1,\dots, x_n)$. The the Koszul cohomology vanishes on dimensions less than $i$, and if $IM\neq M$ then the $i$th dimensional cohomology does not vanish.
I understand every part of his proof, besides the last sentence. I just cannot figure out how he made that jump. I know there are other proofs of this fact, but I am not interested in those.
One knows that $r=\min\{i:H_i({\bf x,y};M)\ne0\}$. Notice that $I=({\bf x,y})$. We also know that $I$ contains a maximal $M$-sequence, namely ${\bf y}$. If $IM\ne M$ then one can apply the Corollary 17.12 and get $s=\min\{i:H_i({\bf x,y};M)\ne0\}$, so $s=r$.
The only thing to show now is $IM\ne M$. Can you do this?